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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Deleted User on April 01, 2013, 07:59:09 pm

Title: Simplify by hand!
Post by: Deleted User on April 01, 2013, 07:59:09 pm

square root (2/square root (3) - 2i) where i is imaginary number

Title: Re: Simplify by hand!
Post by: b^3 on April 01, 2013, 08:05:23 pm

Hint: Try 'rationalising' the denominator first.

Spoiler

$\begin{alignedat}{1}\frac{2}{\sqrt{3}-2i} & =\frac{2}{\sqrt{3}-2i}\times\frac{\sqrt{3}+2i}{\sqrt{3}+2i}<br />\\ & =\frac{2\sqrt{3}+4i}{3-4i^{2}}<br />\\ & =\frac{2\sqrt{3}+4i}{3+4}<br />\\ & =\frac{2\sqrt{3}+4i}{7}<br />\end{alignedat}$

Title: Re: Simplify by hand!
Post by: Phy124 on April 01, 2013, 10:11:47 pm

Quote from: Deleted User on April 01, 2013, 08:53:46 pm

Sorry the -2i part isn't part of the division.
What I meant was sqrt [ (2/sqrt(3)) - 2i ]

We know that:

$\frac{2}{\sqrt{3}}-2i = \frac{4}{\sqrt{3}}\left(\frac{1}{2}-\frac{\sqrt{3}}{2} i\right) = \frac{4}{\sqrt{3}} cis\left(-\frac{\pi}{3}\right)$

Therefore using de Moivre's formula we can show that;

$\sqrt{\frac{2}{\sqrt{3}}-2i}$

$= \left(\frac{4}{\sqrt{3}} cis\left(-\frac{\pi}{3}\right)\right)^{\frac{1}{2}}$

$= \left(\frac{4}{\sqrt{3}}\right)^{\frac{1}{2}}cis\left(-\frac{\pi}{6}\right)$

$= \frac{2}{3^{\frac{1}{4}}} \ cis\left(-\frac{\pi}{6}\right)$

$= \frac{2}{3^{\frac{1}{4}}}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)$

$= 3^{\frac{1}{4}}-\frac{1}{3^{\frac{1}{4}}}i$

Title: Re: Simplify by hand!
Post by: silverpixeli on April 01, 2013, 10:20:19 pm

Am I wrong in saying that (some imaginary number)^1/2 should have two solutions? Or do we only take the principal solution in this case because the original question asked for the sqrt() which is the principal square root?

Title: Re: Simplify by hand!
Post by: Phy124 on April 01, 2013, 11:18:23 pm

Quote from: silverpixeli on April 01, 2013, 10:20:19 pm

Am I wrong in saying that (some imaginary number)^1/2 should have two solutions? Or do we only take the principal solution in this case because the original question asked for the sqrt() which is the principal square root?

I only took the principal root, but I wasn't entirely sure whether I should have given the second as well. Having not done spesh, I'm not sure whether it assumed you list all roots or list the principal and list the others when asked to do so.

I think that it is the latter but on second thought, it could be the former, where you list all solutions of the domain [-pi,pi], in which case the second root will be the negative of the original, i.e.

$-3^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}}i$

tl;dr: I can't answer with confidence because I didn't do the subject, best to ask your teacher :)

edit: thanks for clearing that up abcdqd :)

Title: Re: Simplify by hand!
Post by: abcdqd on April 01, 2013, 11:21:13 pm

if the question asks for the sqrt, you have to reject the negative answer
if it just says find the roots of, you have to list all of them

Title: Re: Simplify by hand!
Post by: Deleted User on April 02, 2013, 09:59:13 am

Thank you!

ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Deleted User on April 01, 2013, 07:59:09 pm